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(C) Basic strength is inversely proportional to the stability of the conjugate base (anion).$1$. $CH_3 - CH_2 - CO_2^-$: The negative charge is deloca

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$D > B > A > C$

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(C) Basic strength is inversely proportional to the stability of the conjugate base (anion).$1$. $CH_3 - CH_2 - CO_2^-$: The negative charge is delocalized over two oxygen atoms via resonance,making it the most stable and least basic.$2$. $CH_3 - CH_2 - C \equiv C^-$: The negative charge is on an $sp$-hybridized carbon atom. Due to high electronegativity of $sp$ carbon,it is relatively stable.$3$. $CH_3 - CH_2 - S^-$: The negative charge is on a large sulfur atom. Due to the large size of $S$,the charge is dispersed,making it more stable than the oxygen anion.$4$. $CH_3 - CH_2 - O^-$: The negative charge is on a small,highly electronegative oxygen atom with no resonance stabilization,making it the least stable and most basic.Comparing the stability of the conjugate acids: $CH_3-CH_2-OH$ $(pKa \approx 16)$ Correct order of basicity: $D > B > A > C$.

List-$I$ (Acid)	List-$II$ $(K_a)$
$A$. Phenol	$I$. $1 \times 10^{-13}$
$B$. Benzoic acid	$II$. $3.0 \times 10^{-8}$
$C$. $HClO$	$III$. $1.0 \times 10^{-10}$
$D$. $CH_3COOH$	$IV$. $6.5 \times 10^{-5}$
	$V$. $1.75 \times 10^{-5}$

Rank the following in decreasing order of basic strength:
$A. CH_3 - CH_2 - C \equiv C^-$
$B. CH_3 - CH_2 - S^-$
$C. CH_3 - CH_2 - CO_2^-$
$D. CH_3 - CH_2 - O^-$

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Rank the following in decreasing order of basic strength:$A. CH_3 - CH_2 - C \equiv C^-$$B. CH_3 - CH_2 - S^-$$C. CH_3 - CH_2 - CO_2^-$$D. CH_3 - CH_2 - O^-$